Solution to Legendre's Differential Equation

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Theorem

The solution of Legendre's differential equation:

$\displaystyle \left({1 - x^2}\right) \frac{\mathrm d^2 y} {\mathrm d x^2} - 2 x \frac{\mathrm d y} {\mathrm d x} + p \left({p + 1}\right) y = 0$

can be obtained by Power Series Solution Method.


Proof

Let:

$\displaystyle y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$

such that:

$a_0 \ne 0$

Differentating with respect to $x$:

$\displaystyle \dot y = \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) x^{k - n - 1}$
$\displaystyle \ddot y = \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2}$


Substituting in the original equation:

\(\displaystyle \left({1 - x^2}\right) \ddot y - 2 x \dot y + p \left({p + 1}\right) y\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({1 - x^2}\right) \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2} - 2 x \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) x^{k - n - 1} + p \left({p + 1}\right) \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2} - x^2 \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2}\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} - 2 x \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) x^{k - n - 1} + p \left({p + 1}\right) \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


The summations are dependent upon $n$ and not $x$.

Therefore it is a valid operation to multiply the $x$'s into the summations, thus:

\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2}\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} - \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n} - 2 \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) x^{k - n} + p \left({p + 1}\right) \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({k - n}\right) \left({k - n - 1}\right) x^{k - n - 2}\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n - 1}\right) - 2 \left({k - n}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


Increasing the summation variable $n$ to $n + 2$:

\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \left({k - n + 2}\right) \left({k - n + 1}\right) x^{k - n}\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


Taking the first 2 terms of the second summation out:

\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \left({k - n + 2}\right) \left({k - n + 1}\right) x^{k - n}\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} + a_0 x^k \left({p \left({p + 1}\right) - k \left({k + 1}\right)}\right) + a_1 x^{k - 1} \left({p \left({p + 1}\right) - k \left({k - 1}\right)}\right)\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} + \sum_{n \mathop = 2}^\infty a_{n - 2} x^{k - n} + a_n \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a_0 x^k \left({p \left({p + 1}\right) - k \left({k + 1}\right)}\right) + a_1 x^{k - 1} \left({p \left({p + 1}\right) - k \left({k - 1}\right)}\right)\) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle {} + \sum_{n \mathop = 2}^\infty x^{k - n} \left({a_{n - 2} \left({k - n - 2}\right) \left({k - n + 1}\right) + a_n \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


Equating each term to $0$:

\((1):\quad\) \(\displaystyle a_0 x^k \left({p \left({p + 1}\right) - k \left({k + 1}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\((2):\quad\) \(\displaystyle a_1 x^{k - 1} \left({p \left({p + 1}\right) - k \left({k - 1}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\((3):\quad\) \(\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \left({a_{n - 2} \left({k - n - 2}\right) \left({k - n + 1}\right) + a_n \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


Take equation $(1)$:

$a_0 x^k \left({p \left({p + 1}\right) - k \left({k + 1}\right)}\right) = 0$

It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$.

Thus:

\(\displaystyle p \left({p + 1}\right) - k \left({k + 1}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle p^2 - k^2 + p - k\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({p - k}\right) \left({p + k}\right) + \left({p - k}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({p - k}\right) \left({p + k + 1}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle p - k\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\, \displaystyle \lor \, \) \(\displaystyle p + k + 1\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle k\) \(=\) \(\displaystyle p\) $\quad$ $\quad$
\(\, \displaystyle \lor \, \) \(\displaystyle k\) \(=\) \(\displaystyle p - 1\) $\quad$ $\quad$


Take equation $(2)$:

$a_1 x^{k - 1} \left({p \left({p + 1}\right) - k \left({k - 1}\right)}\right) = 0$

As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$.

Thus:

\(\displaystyle a_1 \left({p \left({p + 1}\right) - k \left({k - 1}\right)}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle a_1 k\) \(=\) \(\displaystyle 0\) $\quad$ substituting the value of $k$ from $(1)$ $\quad$
\(\displaystyle a_1\) \(=\) \(\displaystyle 0\) $\quad$ as $2 k \ne 0$ $\quad$


Take equation $(3)$:

$\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \left({a_{n - 2} \left({k - n - 2}\right) \left({k - n + 1}\right) + a_n \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)}\right) = 0$

As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.

Thus:

\(\displaystyle 0\) \(=\) \(\displaystyle a_{n - 2} \left({k - n - 2}\right) \left({k - n + 1}\right) + a_n \left({p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\left({k - n + 2}\right) \left({k - n + 1}\right)} {p \left({p + 1}\right) - \left({k - n}\right) \left({k - n + 1}\right)} a_{n - 2}\) $\quad$ $\quad$


Since Legendre's differential equation is a second order ODE, it has two independent solutions.


Solution 1 (for $k = p$)

\(\displaystyle y\) \(=\) \(\displaystyle x^p \sum_{n \mathop = 0}^\infty a_n x^{-n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x^k \left({a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\right)\) $\quad$ $\quad$


\(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\left({p - n + 2}\right) \left({p - n + 1}\right)} {p \left({p + 1}\right) - \left({p - n}\right) \left({p - n + 1}\right)} a_{n - 2}\) $\quad$ from equation $(4)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\left({p - n + 2}\right) \left({p - n + 1}\right) } {n \left({2 p - n + 1}\right)} a_{n - 2}\) $\quad$ $\quad$


Thus:

\(\displaystyle a_2\) \(=\) \(\displaystyle -\frac {p \left({p - 1}\right)} {2 \left({2 p - 1}\right)} a_0\) $\quad$ for $n = 2$ $\quad$
\(\displaystyle a_4\) \(=\) \(\displaystyle -\frac {\left({p - 2}\right) \left({p - 3}\right)} {4 \left({2 p - 3}\right)} a_2\) $\quad$ for $n = 4$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {p \left({p - 1}\right) \left({p - 2}\right) \left({p - 3}\right)} {2 \cdot 4 \left({2 p - 1}\right) \left({2 p - 3}\right)} a_0\) $\quad$ $\quad$
\(\displaystyle a_6\) \(=\) \(\displaystyle -\frac {\left({p - 4}\right) \left({p - 5}\right)} {6 \left({2 p - 5}\right)} a_4\) $\quad$ for $n = 6$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac {p \left({p - 1}\right) \left({p - 2}\right) \left({p - 3}\right) \left({p - 4}\right) \left({p - 5}\right) } {2 \cdot 4 \cdot 6 \left({2 p - 1}\right) \left({2 p - 3}\right) \left({2 p - 5}\right)} a_0\) $\quad$ $\quad$


Similarly:

\(\displaystyle a_{2 n}\) \(=\) \(\displaystyle \left({-1}\right)^n \frac {p \left({p - 1}\right) \left({p - 2}\right) \left({p - 3}\right) \dotsm \left({p - 2 n - 1}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p - 1}\right) \left({2 p - 3}\right) \left({2 p - 3}\right) \dotsm \left({2 p - 2 n + 1}\right)} a_0\) $\quad$ $\quad$


Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$


Substituting for $a_n$:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^k \left({1 + \frac {-p \left({p - 1}\right)} {2 \left({2 p - 1}\right)} x^{-2} + \frac {p \left({p - 1}\right) \left({p - 2}\right) \left({p - 3}\right)} {2 \cdot 4 \left({2 p - 1}\right) \left({2 p - 3}\right)} x^{-4} }\right.\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\left({-1}\right)^n \frac {-p \left({p - 1}\right) \left({p - 2}\right) \dotsm \left({p - 2 n + 1}\right)} {\left({2 \cdot 4 \cdot 6 \dots 2 n}\right) \left({2 p - 1}\right) \left({2 p - 3}\right) \dotsm \left({p - 2 n + 1}\right)} x^{-2 n} + \dotsb}\right)\) $\quad$ $\quad$

$\Box$


Solution 2 (for $k = p - 1$)

\(\displaystyle y\) \(=\) \(\displaystyle x^{-p - 1} \sum_{n \mathop = 0}^\infty a_n x^{-n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x^{-k - 1} \left({a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\right)\) $\quad$ $\quad$



\(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\left({-p - 1 - n + 2}\right) \left({-p - 1 - n + 1}\right)} {p \left({p + 1}\right) - \left({- p - 1 - n}\right) \left({-p - n}\right)} a_{n - 2}\) $\quad$ from equation $(4)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({p + n - 1}\right) \left({p + n}\right)} {n \left({2 p + n + 1}\right)} a_{n - 2}\) $\quad$ $\quad$


Thus:

\(\displaystyle a_2\) \(=\) \(\displaystyle \frac {\left({p + 1}\right) \left({p + 2}\right)} {2 \left({2 p + 3}\right)} a_0\) $\quad$ for $n = 2$ $\quad$
\(\displaystyle a_4\) \(=\) \(\displaystyle \frac {\left({p + 3}\right) \left({p + 4}\right)} {4 \left({2 p + 5}\right)} a_2\) $\quad$ for $n = 4$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \left({p + 4}\right)} {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} a_0\) $\quad$ $\quad$
\(\displaystyle a_6\) \(=\) \(\displaystyle \frac {\left({p + 5}\right) \left({p + 6}\right)} {6 \left({2 p + 7}\right)} a_4\) $\quad$ for $n = 6$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \left({p + 4}\right) \left({p + 5}\right) \left({p + 6}\right)} {2 \cdot 4 \cdot 6 \left({2 p + 3}\right) \left({2 p + 5}\right) \left({2 p + 7}\right)} a_0\) $\quad$ $\quad$


Similarly:

\(\displaystyle a_{2 n}\) \(=\) \(\displaystyle \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \left({2 p + 7}\right) \dotsm \left({2 p + 2 n + 1}\right)} a_0\) $\quad$ $\quad$


Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$


Substituting for $a_n$:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\left({p + 1}\right) \left({p + 2}\right)} {2 \left({2 p + 3}\right)} x^{-2} + \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \left({p + 4}\right)} {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} x^{-4} }\right.\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\left({-1}\right)^n \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \dotsm \left({p - 2 n}\right)} x^{-2 n} + \dotsb}\right)\) $\quad$ $\quad$

$\Box$


In summary, the two solutions are:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^k \left({1 + \frac {-p \left({p - 1}\right)} {2 \left({2 p - 1}\right)} x^{-2} + \frac {p \left({p - 1}\right) \left({p - 2}\right) \left({p - 3}\right)} {2 \cdot 4 \left({2 p - 1}\right) \left({2 p - 3}\right)} x^{-4} }\right.\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\left({-1}\right)^n \frac {-p \left({p - 1}\right) \left({p - 2}\right) \dotsm \left({p - 2 n + 1}\right)} {\left({2 \cdot 4 \cdot 6 \dots 2 n}\right) \left({2 p - 1}\right) \left({2 p - 3}\right) \dotsm \left({p - 2 n + 1}\right)} x^{-2 n} + \dotsb}\right)\) $\quad$ $\quad$


and:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\left({p + 1}\right) \left({p + 2}\right)} {2 \left({2 p + 3}\right)} x^{-2} + \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \left({p + 4}\right)} {2 \cdot 4 \left({2 p + 3}\right) \left({2 p + 5}\right)} x^{-4} }\right.\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\left({-1}\right)^n \frac {\left({p + 1}\right) \left({p + 2}\right) \left({p + 3}\right) \dotsm \left({p + 2 n}\right)} {\left({2 \cdot 4 \cdot 6 \dotsm 2 n}\right) \left({2 p + 3}\right) \left({2 p + 5}\right) \dotsm \left({p - 2 n}\right)} x^{-2 n} + \dotsb}\right)\) $\quad$ $\quad$


$\blacksquare$


Also see

The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.


Source of Name

This entry was named for Adrien-Marie Legendre.